College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 14 - Problems - Page 533: 3


An internal energy of $4900~J$ is generated.

Work Step by Step

The increase in internal energy is equal to the initial kinetic energy of the bullet: $\Delta E = \frac{1}{2}mv^2$ $\Delta E = \frac{1}{2}(0.020~kg)(7.00\times 10^2~m/s)^2$ $\Delta E = 4900~J$ An internal energy of $4900~J$ is generated.
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